Acids, Bases, and Aqueous Equilibria

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Acids, Bases, and Aqueous Equilibria
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Nature of Acids and Bases

• Acids--
• Sour taste, corrosive to metals
• Bases--
• Bitter taste, feel slippery, corrosive to fat
• Dont use these to identify acids/bases in lab!

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Definitions of Acids and Bases

• Arrhenius Concept Acids produce H in solution,
  bases produce OH? ion.
• Brønsted-Lowry Acids are H donors, bases are
  proton acceptors.
• HCl H2O ? Cl? H3O
• acid base

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Conjugate Acid/Base Pairs

• HA(aq) H2O(l) ? H3O(aq) A?(aq)
• conj conj
• acid 1 base 2 acid 2
  base 1
• conjugate base everything that remains of the
  acid molecule after a proton is lost.
• conjugate acid formed when the proton is
  transferred to the base.

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Acid Dissociation Constant (Ka)

• An equilibrium exists in water solutions of acids
  
• HA(aq) H2O(l) ? H3O(aq) A?(aq)
• or HA(aq) ? H (aq) A- (aq)

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Example 14.1

• Give dissociation reactions for these HCl,
  HC2H3O2, NH4, C6H5NH3
• HCl ? H Cl-
• HC2H3O2 ? H C2H3O2-
• NH4 ? H NH3
• C6H5NH3 ? H C6H5NH2

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Acid Strength
Strong Acid

• Its equilibrium position lies far to the right.
  (HNO3, HCl, HBr, HI, HClO4, H2SO4) Ka gtgt 1
  these are the ONLY strong acids
• Yields a weak conjugate base. (NO3?, or others
  from above acids)
• H2SO4 is only strong in its 1st H

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Acid Strength(continued)
Weak Acid

• Its equilibrium lies far to the left. (CH3COOH,
  and other organic acids) Ka ltlt 1
• Yields a much stronger (it is relatively strong)
  conjugate base than water. (CH3COO?)

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Types of Acids

• Binary Acids Hydrogen bonded to elements other
  than oxygen, which has acid characteristics HCl,
  HCN, H2S
• Oxyacids Hydrogen bonded to a polyatomic ion
  containing oxygenH2CO3, H3PO4
• Organic Acidscontain the carboxyl group,
• OH
• -CO which are all weak acids

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Strong Acids
Weak Acids
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Strong
Weak
Weak
Strong
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Example 14.2

• From the previous slide, arrange these bases from
  weak to stronger H2O, F-, Cl-, NO2-, CN-
• Cl- is from strong acid, as is H2O (from H3O),
  so both are very weak. CN- is from the weakest
  acid and is therefore the strongest. HF is a
  stonger acid than HNO2, so NO2- is stronger than
  F-, so the ranking from weak to strong is
• Cl- lt H2O lt F- lt NO2- lt CN-

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Water as an Acid and a Base

• Water is amphoteric (it can behave either as an
  acid or a base).
• H2O H2O ? H3O OH?
• 
  conj conj
• acid base acid base
• Kw 1 ? 10?14 at 25C H OH-
• Must always be a balance between H and OH-

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Example 14.3

• Calculate H and OH- in these solutions
• a. 1.0 x 10-5 M OH-
• b. 1.0 x 10-7 M OH-
• c. 10.0 M H
• OH- 1.0 x 10-5 M H Kw / OH- 1 x
  10-14/ 1.0 x 10-5 1.0 x 10-9 M
• OH- 1.0 x 10-7 M H Kw / OH- 1 x
  10-14/ 1.0 x 10-7 1.0 x 10-7 M
• H 10.0 M OH- Kw / H 1 x
  10-14/10.0 1.0 x 10-15 M

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The pH Scale

• There is a more convenient way to indicate H
• pH ? ?logH
• pH in water normally ranges from 0 to 14, but can
  extend to negative or gt14 values
• Kw 1.00 ? 10?14 H OH?
• pKw 14.00 pH pOH
• As pH rises, pOH falls (sum 14.00).

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Example 14.5

• Calculate pH and pOH for each of these
• solutions a. 1.0 x 10-3 M OH- b. 1.0 M H
• -log OH- 3 pOH pH 14 pOH 11
• b. -log H 0 pH pOH 14 pH 14

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Example 14.6 7

• If pH of blood is 7.41, find pOH, H, and OH-
• pOH 14-pH 6.59 H 10-pH 10-7.41 3.9 x
  10-8 M
• OH- 10-6.59 2.6 x 10-7 M
• Calculate pH for 0.10 M HNO3 and 1 x 10-10 M HCl
• Since both are strong acids, they are totally
  dissociated, and H acid strength of the
  major species. pH of HNO3 is therefore
  log(0.10) 1. But the HCl solution is so
  dilute that the water provides most of the H,
  and so the pH7.

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HOMEWORK 14a

• p. 673ff
• 27, 28, 29, 30, 31, 32, 33, 41, 42

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Solving Weak Acid Equilibrium Problems

• List major species in solution.
• Choose species that can produce H and write
  reactions.
• Based on K values, decide on dominant
  equilibrium.
• Write equilibrium expression for dominant
  equilibrium.
• List initial concentrations in dominant
  equilibrium. (I)

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Solving Weak Acid Equilibrium Problems (continued)

• Define change at equilibrium (as x). (C)
• Write equilibrium concentrations in terms of x.
  (E)
• Substitute equilibrium concentrations into
  equilibrium expression.
• Solve for x the easy way.
• Verify assumptions using 5 rule.
• Calculate H and pH.

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Example 14.8
x2 3.5 x 10-8(0.1) 3.5 x 10-9 x 5.9 x 10-5 M
H pH -log(5.9 x 10-5) 4.23

• Calculate the pH of a 0.100 M solution of HOCl
  (Ka3.5 x 10-8)
• HOCl H OCl-
• I 0.1 0 0
• C -x x x
• E 0.1-x x x

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Percent Dissociation (Ionization)
Calculate this from H
This becomes greater as the acid concentration
becomes more diluteExample 14.10
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Example 14.10

• Calculate percent dissociation for a. 1.00 M
  HC2H3O2 and b. 0.100 M HC2H3O2 Ka 1.8 x
  10-5
• Acid H A- b. Acid H
  A-
• I 1.00 0 0 0.100
  0 0
• C -x x x -x
  x x
• E 1-x x x 0.1-x
  x x
• 1.8 x 10-5 x2 / 1 1.8 x 10-5 x2 / 0.1
• x H 4.2 x 10-3 M x 1.3 x 10-3 M
• diss 4.2 x 10-3 / 1.0 diss 1.3 x 10-3
  / 0.100
• 0.42 1.3

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Homework 14b

• p. 675ff
• 55, 59, 60, 63, 64, 68

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Bases

• Strong and weak are used in the same sense
  for bases as for acids.
• strong complete dissociation (hydroxide ion
  supplied to solution) Most common are metal
  hydroxides. Kb is very large.
• NaOH(s) ? Na(aq) OH?(aq)

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Example 14.12

• Calculate pH of 5.0 x 10-2 M NaOH
• Strong base means OH- NaOH 0.05 M
• pOH -log(0.05) 1.30
• pH 14- pOH 12.70

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Bases(continued)

• weak very little dissociation (or reaction with
  water) Usually contain an -NHn group
• H3CNH2(aq) H2O(l) ? H3CNH3(aq) OH?(aq)
• Kb for weak bases is usually very small lt 10-3

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Calcualtions Involving Weak Bases
Use the same ICE method as with weak acids Notice
that x will equal OH- rather than
H Calculate pOH and from that calculate pH
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Example 14.13

• Calculate the pH of a 15.0 M solution of NH3
• Kb 1.8 x 10-5 NH3 NH4 OH-
• I 15.0 0 0
• C -x x x
• E 15-x x x
• 1.8 x 10-5 x2 / 15 x2 1.8 x 10-5 (15)2.7 x
  10-4
• x 1.6 x 10-2 M OH- pOH 1.80
• pH 14-pOH12.2

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Polyprotic Acids

• . . . can furnish more than one proton (H) to
  the solution. Each one comes off separately.

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Acid-Base Properties of Salts
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Relationship of Ka to Kb

• For acidic/basic salts, Ka or Kb must be
  calculated from the parent acid/base value
• Kb Kw Ka Kw
• Ka Kb
• Example
• Ka of HC2H3O2 1.8 x 10-5
• Kb 1 x 10-14 / 1.8 x 10-5 5.6 x 10-10

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Structure and Acid-Base Properties

• Two factors for acidity in binary compounds
• Bond Polarity (high is good)
• Bond Strength (low is good)

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Oxides

• Acidic Oxides (Acid Anhydrides)
• O?X bond is strong and covalent.
• SO2, NO2, CrO3
• Basic Oxides (Basic Anhydrides)
• O?X bond is ionic.
• K2O, CaO

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Lewis Acids and Bases

• Lewis Acid electron pair acceptor
• Lewis Base electron pair donor

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Homework 14c

• p. 675ff
• 62, 66, 71, 72, 77, 80, 84, 88, 94

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Buffers, Titrations, and Aqueous Equilibria
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Common Ion Effect

• The shift in equilibrium that occurs because of
  the addition of an ion already involved in the
  equilibrium reaction. (Le Chateliers principle)
• AgCl(s) ? Ag(aq) Cl?(aq)
• This affects the concentrations of other ions,
  notably H

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Calculations with ICE
What is the H and ionization of 1.0 M HF
mixed with 1.0 M NaF. (1.0 M HF alone, H2.7 x
10-2, ion2.7) HF(aq) ltgt H (aq) F-
(aq) I 1.0 0 1.0 C -x x x E
1.0-x x 1.0 x Ka 7.2 x 10-4 x
(1.0x) / (1.0 - x) x/1 xH 7.2 x 10-4
ion 7.2 x 10-4 /1.0 .072
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A Buffered Solution

• . . . resists change in its pH when either H or
  OH? are added.
• 1.0 L of 0.50 M H3CCOOH
• 0.50 M H3CCOONa
• pH 4.74
• Adding 0.010 mol solid NaOH raises the pH of the
  solution to 4.76, a very minor change.

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Key Points on Buffered Solutions

• 1. They are weak acids or bases containing a
  common ion.
• 2. After addition of strong acid or base, deal
  with stoichiometry first, then equilibrium.

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Demonstration of Buffer Action
A buffered solution of 1.0 L of 0.5 M HC2H3O2
Ka1.8 x 10-5 and 0.5 M NaC2H3O2 has 0.01 mol of
solid NaOH added. What is the new pH? HC2H3O2
(aq)ltgt C2H3O2-(aq) H (aq) I
0.5 0.5 0 C -x x x E 0.5
-x 0.5 x x Ka1.8 x 10-5 0.5(x) / 0.5
x 1.8 x 10-5 pH 4.74
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When OH- is added, it takes away an equal amount
of H ions, and will affect also the acid
concentration by the same amount. It will also
add to the common ion. HC2H3O2 (aq)ltgt
C2H3O2-(aq) H (aq) I 0.5 0.5 1.8 x
10-5 S -0.01 0.01 - 1.8 x 10-5 due
to adding base I 0.49 0.51 0 C -x x x E
0.49-x 0.51 x x Ka 1.8 x 10-5 x (.51)/ .49
x 1.73 x 10-5 pH 4.76 Practically no
change in pH after base is added.
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Henderson-Hasselbalch Equation

• Useful for calculating pH when the
  A?/HA ratios are known.

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Base Buffers With H/H Equation
pOH pKb log ( HB / B ) pKb log
(acid/ base) Base buffers can be calculated
in similar fashion to acid buffers. pH 14 - pOH
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Buffered Solution Characteristics

• Buffers contain relatively large amounts of weak
  acid and corresponding base.
• Added H reacts to completion with the weak base.
• Added OH? reacts to completion with the weak
  acid.
• The pH is determined by the ratio of the
  concentrations of the weak acid and weak base.

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H/H Calculations
From the previous example, HC2H3O2 .49 M
after the addition of base, and C2H3O2- .51
M. Using the H/H equation, pH pKa log
(.51)/(.49) 4.74 0.02 4.76 same as
using ICE, but easier A weak base buffer is made
with 0.25 M NH3 and 0.40 M NH4Cl. What is the pH
of this buffer? Kb 1.8 x 10-5 NH3 (aq) H2O
(l) ltgt NH4 (aq) OH- (aq) pOH pKb log
(NH4 / NH3) 4.74 log (.40/.25) 4.94 pH
14- 4.94 9.06
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H/H Calculations
The weak base buffer with pH of 9.06 from the
previous example has 0.10 mol of HCl added to it.
What is the new pH? NH3 (aq) H2O (l)
ltgt NH4 (aq) OH- (aq) I .25 .40
1.15 x 10-5 S - .10 .10 - 1.15
x 10-5 due to acid I .15 .50
0 Using H/H, pOH 4.74 log (.5)/(.15)
5.26 pH 14 - 5.26 8.74
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Buffering Capacity

• . . . represents the amount of H or OH? the
  buffer can absorb without a significant change in
  pH.

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Homework 15a

• p. 741ff 21, 23d, 24d, 37, 38, 43

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Titration (pH) Curve

• A plot of pH of the solution being analyzed as a
  function of the amount of titrant added.
• Equivalence (stoichiometric) point Enough
  titrant has been added to react exactly with the
  solution being analyzed.

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Strong Acid/Base Titration
In titrations, it is easier to calculate
millimoles(mmol), which would be Molarity x mL,
in stoichiometry. What would be the pH of 50.0 mL
of a 0.2 M solution of HCl after 20.0 mL of 0.1 M
NaOH have been added? HCl---50.0 x 0.2 10.0 mmol
NaOH--- 20.0 x 0.1 2.0 mmol H
OH- ----gt H2O volumes must be added I
10mmol 2 mmol 50 20 70 S -2
mmol -2mmol H 8 mmol/ 70 mL 0.11 M
End 8mmol 0 mmol pH 0.95
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Weak Acid - Strong Base Titration

• Step 1 - A stoichiometry problem - reaction is
  assumed to run to completion - then determine
  remaining species.
• Step 2 - An equilibrium problem - determine
  position of weak acid equilibrium and calculate
  pH.

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Weak Acid/Strong Base
First do a stoichiometry, then equilibrium using
H/H equation What is the pH of 50.0 mL of a
0.100M HCN solution after 8.00 mL of 0.100 M NaOH
has been added? Ka 6.2 x 10-10 HCN OH-
-----gt H2O CN- Volume I 5 mmol 0.8
mmol 0 mmol 50 8 58 S - 0.8 mmol
-0.8 mmol 0.8 mmol I 4.2 mmol /58mL
0.8 mmol /58 mL 0.072M 0.0138 M pH pKa
log (0.0138/0.072) 8.49
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Acid-Base Indicator

• . . . marks the end point of a titration by
  changing color.
• The equivalence point is not necessarily the same
  as the end point.

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pH
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0
1
2
3
4
5
6
7
8
9
10
11
12
13
Crystal Violet
Cresol Red
Thymol Blue
Erythrosin B
2,4-Dinitrophenol
Bromphenol Blue
Methyl Orange
Bromcresol Green
Methyl Red
Eriochrome Black T
Bromcresol Purple
Alizarin
Bromthymol Blue
Phenol Red
o
-Cresolphthalein
Phenolphthalein
Thymolphthalein
Alizarin Yellow R
The pH ranges shown are approximate. Specific
transition ranges depend on the indicator solvent
chosen.
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Homework 15b

• p. 741 51, 57, 71 (for 57 only)